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Abstract <p> Let <InlineEquation ID="IEq1"> <EquationSource Format="TEX">\(\mathcal A\)</EquationSource> </InlineEquation> be a complex unital commutative Banach algebra. Let <InlineEquation ID="IEq2"> <EquationSource Format="TEX">\(\varphi\colon \mathcal A\to \mathbb C\)</EquationSource> </InlineEquation> be a map such that for <InlineEquation ID="IEq3"> <EquationSource Format="TEX">\(x,y\in\mathcal A\)</EquationSource> </InlineEquation>, <InlineEquation ID="IEq4"> <EquationSource Format="TEX">\(\varphi(x)-\varphi(y)\in\sigma_\varepsilon(x-y)\)</EquationSource> </InlineEquation> and <InlineEquation ID="IEq5"> <EquationSource Format="TEX">\(\varphi\)</EquationSource> </InlineEquation> has <InlineEquation ID="IEq6"> <EquationSource Format="TEX">\(\mathbb C\)</EquationSource> </InlineEquation>-linear differentials almost everywhere. Then <InlineEquation ID="IEq7"> <EquationSource Format="TEX">\(\varphi\)</EquationSource> </InlineEquation> is approximately multiplicative. A similar conclusion is reached by replacing the differential condition with comparable assumptions on the map. This result is similar to the Kowalski–Słodkowski theorem. Analogous versions of it are also discussed for the exponential spectrum and for a particular class of the Ransford spectrum. </p>

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Kowalski–Słodkowski Theorem for Spectrum Variants

  • 期刊论文

  • 发表日期：

    2026年3月31日

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  • Gayathri Sugirtha,
  • Daniel Sukumar

摘要

Abstract

Let \(\mathcal A\) be a complex unital commutative Banach algebra. Let \(\varphi\colon \mathcal A\to \mathbb C\) be a map such that for \(x,y\in\mathcal A\) , \(\varphi(x)-\varphi(y)\in\sigma_\varepsilon(x-y)\) and \(\varphi\) has \(\mathbb C\) -linear differentials almost everywhere. Then \(\varphi\) is approximately multiplicative. A similar conclusion is reached by replacing the differential condition with comparable assumptions on the map. This result is similar to the Kowalski–Słodkowski theorem. Analogous versions of it are also discussed for the exponential spectrum and for a particular class of the Ransford spectrum.